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Simple algebra, i don't know how to do!

  1. Oct 4, 2005 #1
    well 'm finding the unit tangent bector and normal vector and then finding its curve, thats not bad but i'm hung up on some algebra..
    I have sqrt(4t^2 + 4 + 1/t^2);
    can that equal...
    2t*2*1/t = 4?

    Also...
    sqrt(4t^2 + 25t^8)
    Will that simpify too:
    10t^5?
     
  2. jcsd
  3. Oct 4, 2005 #2

    Doc Al

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    Staff: Mentor

    Why don't you test them with a few values of t? (Try t = 1.)
     
  4. Oct 4, 2005 #3
    Yeah that is so not working, so i know its wrong and yet i'm still stuck. Who said 8th grade math was required for calc III? :uhh:
     
  5. Oct 5, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Hopefully, you learned long ago that, in general, [itex]\sqrt{a^2+ b^2}\ne a+ b[/itex].
    What is (2t+ 1/t)2? Do you see how that answers your question?

    "Who said 8th grade math was required for calc III? "

    Pretty much every teacher of 8th grade math or calc III!
     
  6. Oct 5, 2005 #5

    Diane_

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    Homework Helper

    I used to tell my calculus 1 students that everything they had ever learned in math, from kindergarten on, would show up sometime during the year, and anything they had learned imperfectly would grow teeth and bite them. No one ever accused me of exaggerating by the end of the year.
     
  7. Oct 5, 2005 #6
    Try to square your answer :
    [tex](2t * 2 *\frac{1}{t})^2[/tex]

    And hopefully you'll see that it's not equal to what you originally had, which is written like:
    [tex]\sqrt{4t^2 + 4 + \frac{1}{t^2}}[/tex]

    Same for:
    [tex]\sqrt{4t^2 + 25t^8}[/tex],
    that will not equal
    [tex]10t^5[/tex]
    Can you tell us how you got that so at least one of our many brilliant helpers can assist you on your algebra, and hopefully overcoming your weakness.
     
    Last edited: Oct 5, 2005
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